Bounding the max number of points determining 2 distances

Question

My professor discussed a generalization of the unit distance problem:

Bound $m(n)$, the maximum number of points $x_1,\dots,x_m\in \mathbb{R}^n$ s.t. $|\{||x_i-x_j||:i,j\in[m]\}|=2$.

She mentioned that an easy lower bound is $m(n)\geq n(n+1)/2$. I've been trying to think of configurations of points to realize this bound, but I've had no success. I do notice it's equal to ${n+1 \choose n}=1+\cdots+n$. Maybe this is relevant?

Answer 1

If we take $x_{\{i,j\}}\in\mathbb{R}^n$ as the vector whose only non-zero components are the $i$-th coordinate and the $j$-th coordinate, equal to one, then for any $A,B\subset\{1,\ldots,n\}$ such that $A\neq B, |A|=|B|=2$ we have that the distance between $x_A$ and $x_B$ is either $2$ or $\sqrt{2}$.

This shows $m(n)\geq \frac{n(n-1)}{2}$ and I guess your professor was referring to some little improvement of this idea.